The general case:

\begin{matrix} 1^{m} + 2^{m} + \dots + n^{m} \\ = {(1^{m} e^{t} + 2^{m} e^{2 t} + \dots n^{m} e^{nt}) |}_{t = 0} \\ = {\frac{d^{m}}{{dt}^{m}} (1 + e^{t} + e^{2 t} + \dots + e^{nt}) |}_{t = 0} \\ = {\frac{d^{m + 1}}{{dt}^{m + 1}} \frac{t (e^{(n + 1) t} - 1)}{e^{t} - 1} |}_{t = 0} \\ = \frac{1}{m + 1} ({\frac{d^{m + 1}}{{dt}^{m + 1}} \frac{{te}^{(n + 1) t}}{e^{t} - 1} |}_{t = 0} - {\frac{d^{m + 1}}{{dt}^{m + 1}} \frac{t}{e^{t} - 1} |}_{t = 0}) \\ = \frac{1}{m + 1} ({\frac{d^{m + 1}}{{dt}^{m + 1}} \sum_{k = 0}^{\infty} B_{k} (n + 1) \frac{t^{k}}{k!} |}_{t = 0} - {\frac{d^{m + 1}}{{dt}^{m + 1}} \sum_{k = 0}^{\infty} B_{k} \frac{t^{k}}{k!} |}_{t = 0}) \\ = \frac{B_{m + 1} (n + 1) - B_{m + 1}}{m + 1} \end{matrix}

Where

B_{n} (x)

is the Bernoulli polynomial and B_n are the Bernoulli numbers

Conclude that

$1^{m} + 2^{m} + \dots + n^{m} = \frac{B_{m + 1} (n + 1) - B_{m + 1}}{m + 1}$

Amongst other things, all you have to do now is find out what on earth is a Bernoulli polynomial!!